Find the area of a triangle with side lengths 8, 9, and 9.
Solution: We draw the longest altitude of this triangle, which breaks the triangle into two right triangles.   [asy]
size(100);
pair A,B,C,D;
A=(0,0); B=(8,0); C=(4,sqrt(65)); D=(4,0);
draw(A--B--C--cycle); draw(C--D);
draw(rightanglemark(C,D,A,18));
label("8",D,S); label("9",(A+C)/2,W); label("9",(B+C)/2,E);
[/asy] The right triangles are congruent because the original triangle is isosceles; each right triangle has one leg length $8/2=4$ and hypotenuse length 9.  The other leg, which is also the height of the original triangle, has length $\sqrt{9^2-4^2}=\sqrt{65}$.  Thus, the original triangle has base length 8, height $\sqrt{65}$, and area \[\frac{1}{2}(8)(\sqrt{65})=\boxed{4\sqrt{65}}.\]